Note Mat. 35 (2015) no. 1, 1–13. doi:10.1285/i15900932v35n1p1

Postulation of general unions of lines and decorated lines

Edoardo Ballico^{i}

Department of Mathematics, University of Trento ballico@science.unitn.it

Received: 2.11.2013; accepted: 3.6.2014.

Abstract. A +line A ⊂ P^{r}, r ≥ 3, is the scheme A = L ∪ v with L a line and v a tangent
vector of P^{r} supported by a point of L, but not tangent to L. Here we prove that a general
disjoint union of lines and +lines has the expected Hilbert function.

Keywords:postulation, Hilbert function, lines, zero-dimensional schemes

MSC 2000 classification:primary 14N05, secondary 14H99

Introduction

Fix a line L ⊂ P^{r}, r ≥ 2, and P ∈ L. A tangent vector of P^{r} with P
as its support is a zero-dimensional scheme Z ⊂ P^{r} such that deg(Z) = 2 and
Zred= {P }. The tangent vector Z is uniquely determined by P and the line hZi
spanned by Z. Conversely, for each line D ⊂ P^{r} with P ∈ D there is a unique
tangent vector v with v_{red}= P and hvi = D. A +line M ⊂ P^{r} supported by L
and with a nilradical at P is the union v ∪ L of L and a tangent vector v with P
as its support and spanning a line hvi 6= L. The set of all +lines of P^{r}supported
by L and with a nilradical at P is an irreducible variety of dimension r − 1 (the
complement of L in the (r − 1)-dimensional projective space of all lines of P^{r}
containing P ). Hence the set of all +lines of P^{r} supported by L is parametrized
by an irreducible variety of dimension r. Therefore the set of all +lines of P^{r}
is parametrized by an irreducible variety of dimension 2(r − 1) + r = 3r − 1.

Now assume r ≥ 3. For all integers t ≥ 0 and c ≥ 0 let L(r, t, c) be the set of all
disjoint unions X ⊂ P^{r} of t lines and c +lines. If (t, c) 6= (0, 0), then L(r, t, c) is
an irreducible variety of dimension (t + c)(2r − 1) + cr. Fix any X ∈ L(r, t, c)
and any integer k > 0. It is easy to check that h^{0}(OX(k)) = (k + 1)(t + c) + c
and h^{i}(O_{X}(k)) = 0 for all i > 0 (Lemma 2). A closed subscheme E ⊂ P^{r} is
said to have maximal rank if for every integer k > 0 either h^{0}(I_{E}(k)) = 0 or
h^{1}(IE(k)) = 0, i.e. h^{0}(IE(k)) = max{0, ^{r+k}_{r} − h^{0}(OE(k))}.

iThis work is partially supported by MIUR and GNSAGA (INDAM) http://siba-ese.unisalento.it/ c 2015 Universit`a del Salento

Theorem 1. Fix integers r ≥ 3, t ≥ 0 and c ≥ 0 such that (t, c) 6= (0, 0). If r ≥ 4, then assume that the characteristic is zero. Then a general X ∈ L(r, t, c) has maximal rank.

We prove Theorem 1 for r = 3 in arbitrary characteristic, while we assume
characteristic zero if r ≥ 4. We also get intermediate results (e.g. B_{r,k}) that
may be useful as a sample of lemmas which may be proved with +lines. We see
+lines as a tool to prove something involving the Hilbert function of unions of
curves and fat points. For an alternative approach to such disjoint unions, see
Remark 1.

1 Preliminaries

Remark 1. Fix a line L ⊂ P^{n}, n ≥ 2, and a linear system V ⊆ H^{0}(O_{P}^{n}(k)).

Let L^{(1)} be the first infinitesimal neighborhood of L in P^{n}, i.e. the closed sub-
scheme of P^{n} with (IL)^{2} as its ideal sheaf. Let A be any +line with L as its
support. For any closed subscheme B ⊂ P^{n} set V (−B) := {f ∈ V : f_{|B} ≡ 0}.

The +line A gives independent conditions to V with the only restriction of that L is the support of L if either V (L) = {0} or dim(V (−A)) = dim(V (−L)) − 1.

A general +lines with L as its supports does not give independent conditions to
V with the only restriction that L is its support if and only if V (−L) 6= {0} and
V (−L^{(1)}) = V (−L). Now assume dim(V (−L^{(1)})) = dim(V (−L)) − γ for some
γ > 0. The integer γ is the maximal number of tangent vectors v1, . . . , vγ of P^{n}
supported by points of L and imposing independent conditions to V (−L) (with
the restriction that their support is a point of L). So if we only need an integer
t + c ≥ 2, t + c disjoint lines and x ≥ 2 tangent vectors supported by some of
these lines we may decide to put more than one tangent vector on a single line.

Lemma 1. Let X ⊂ P^{r} be a closed subscheme such that the nilradical sheaf
η ⊆ O_{X} is supported by finitely many points. Set Y := X_{red} and fix k ∈ N.

Then:

(1) χ(O_{X}(k)) = χ(O_{Y}(k)) + deg(η);

(2) h^{0}(I_{X}(k)) ≤ h^{0}(I_{Y}(k)) ≤ h^{0}(I_{X}(k)) + deg(η);

(3) h^{1}(I_{Y}(k)) ≤ h^{1}(I_{X}(k)) ≤ h^{1}(I_{Y}(k)) + deg(η);

(4) h^{0}(I_{X}(k)) − h^{1}(I_{X}(k)) = h^{0}(I_{Y}(k)) − h^{1}(I_{Y}(k)) − deg(η).

Proof. By the definition of the reduction of a scheme the sheaf η is the ideal
sheaf of Y in X. We have exact sequence (respectively of O_{X}-sheaves and of
O_{P}^{r}-sheaves):

0 → η → O_{X}(k) → O_{Y}(k) → 0 (1)

0 → I_{X}(k) → I_{Y}(k) → η → 0 (2)
Since η is supported by finitely many points, we have h^{i}(η) = 0 for all i > 0 and
deg(η) = h^{0}(η). Use the cohomology exact sequences of (1) and (2). ^{QED}

Remark 2. Fix integers r ≥ 3, t ≥ 0 and c > 0. Fix A ∈ L(r, t, c), D ∈
L(r, t + c, 0) and set B := A_{red}.

(1) We have B ∈ L(r, t + c, 0). If A is general in L(r, t, c), then B is general in L(r, t + c, 0).

(2) Assume that D is general in L(r, t + c, 0) and fix a decomposition D =
D_{1} t D_{2} with D_{1} ∈ L(r, t, 0) and D_{2} ∈ L(r, c, 0). Let E be a general
element of L(r, 0, c) with Ered= D2. Then D1 is general in L(r, t, 0), D2

is general in L(r, c, 0) and D_{1}∪ E is general in L(r, t, c).

Lemma 1 and Remark 2 give the following result.

Lemma 2. Fix integers r ≥ 3, t ≥ 0 and c > 0. Fix X ∈ L(r, t, c) and
set Y := Xred. We have Y ∈ L(r, t + c, 0). If A is general in L(r, t, c), then
B is general in L(r, t + c, 0). For each integer k > 0 we have h^{1}(O_{X}(k)) = 0,
h^{0}(O_{X}(k)) = (t + c)(k + 1) + c, h^{0}(I_{Y}(k)) − c ≤ h^{0}(I_{X}(k)) ≤ h^{0}(I_{Y}(k)) and
h^{1}(IY(k)) ≤ h^{1}(IX(k)) ≤ h^{1}(IY(k)) + c.

For all integers r ≥ 3 and k ≥ 0 let H_{r,k} denote the following statement:

Assertion H_{r,k}, r ≥ 3, k ≥ 0: Fix (t, c) ∈ N^{2} \ {(0, 0)} and take a
general X ∈ L(r, t, c). If (k + 1)t + (k + 2)c ≥ ^{r+k}_{k} , then h^{0}(IX(k)) = 0. If
(k + 1)t + (k + 2)c ≤ ^{r+k}_{k} , then h^{1}(IX(k)) = 0.

Lemma 3. Fix a general X ∈ L(r, t, c). If 2t + 3c ≤ r + 1, then h^{1}(I_{X}(1)) =
0. If 2t + 3c ≥ r + 1, then h^{0}(I_{X}(1)) = 0.

Proof. Since the case c = 0 is obvious, we may assume c > 0 and use induction
on c. Fix a general Y ∈ L(r, t, c− 1) and write X = Y t A with A a general +line
of P^{r}. If 2t+3(c−1) ≥ r −1, we immediately see that h^{0}(IY ∪Ared(1)) = 0. Hence
h^{0}(I_{X}(1)) = 0. Hence we may assume 2t + 3(c − 1) ≤ r − 2. Let M ⊂ P^{r} be the
(2t + 3c − 4)-dimensional linear subspace spanned by Y . Since A is general, it
spans a plane N such that M ∩ N = ∅. Hence h^{0}(IY ∪A(1)) = h^{0}(IY(1)) − 3 =

r + 1 − 2t − 3c. ^{QED}

Remark 3. Fix an integer r ≥ 3. By the definition of maximal rank and
the irreducibility of each L(r, t, c) Theorem 1 is true for the integer r if and only
if all Hr,k are true. Since Hr,0 is obviously true, to prove Theorem 1 in P^{r} it is
sufficient to prove H_{r,k} for all k > 0. Lemma 3 says that H_{r,1} is true.

Remark 4. Fix integers r ≥ 3 and k > 0 and suppose you want to prove
H_{r,k}. Fix (t, c) ∈ N^{2}\ {(0, 0)}. First assume t > 0 and (k + 1)(t + c) + c < ^{r+k}_{r} .

Hence (k + 1)(t + c) + (c + 1) ≤ ^{r+k}_{r} . Suppose that h^{1}(I_{X}(k)) = 0 for a general
X ∈ L(r, t−1, c+1). Then h^{1}(IY(k)) = 0 for a general Y ∈ L(r, t, c) (Lemma 1).

Now assume c > 0 and (k + 1)(t + c) + c > ^{r+k}_{r} and so (k + 1)(t + c) + (c − 1) ≥

r+k

r . Suppose that h^{0}(IA(k)) = 0 for a general A ∈ L(r, t + 1, c − 1). Then
h^{0}(I_{B}(k)) = 0 for a general B ∈ L(r, t, c). Therefore to prove H_{r,k} it is sufficient
to test all (t, c) such that either (k + 1)(t + c) + c = ^{k+r}_{r}

or t = 0 and
(k + 2)c < ^{k+r}_{r}

or c = 0 and (k + 1)t > ^{r+k}_{r} . We do not need to test
the pairs (t, 0) by [6]. Among the pairs (0, c) with (k + 2)c ≤ ^{r+k}_{r} it is sufficient
to test the ones with ^{r+k}_{r} − k − 1 ≤ (k + 2)c ≤ ^{r+k}_{r} .

For all integers r ≥ 3 and k ≥ 0 define the integers mr,k and nr,k by the relations

(k + 1)m_{r,k}+ n_{r,k}=r + k
k

, 0 ≤ n_{r,k}≤ k (3)

Remark 5. Fix integers r ≥ 3 and k > 0. Since m_{r,k} ≤ k and k(k + 1) ≤

k+3

3 ≤ ^{r+k}_{r} , we get m_{r,k}≥ n_{r,k}.

For all integers r ≥ 3 and k ≥ 0 set ur,k := d ^{r+k}_{r} /(k + 2)e and v_{r,k} :=

(k + 2)u_{r,k}− ^{r+k}_{r} .

Notice that

(k + 2)(u_{r,k}− v_{r,k}) + (k + 1)v_{r,k}=r + k
r

(4)
and that 0 ≤ v_{r,k}≤ k + 1.

For all integers k > 0 let Ar,k denote the following assertion:

Assertion A_{r,k}, k > 0: Let X ⊂ P^{r} be a general union of v_{r,k} lines and
ur,k− v_{r,k} +lines. Then h^{0}(IX(k)) = 0.

2 The proof in P^{3}

In this section we prove the case r = 3 of Theorem 1.

Lemma 4. Fix integers a ≥ 0, b ≥ 0, y ≥ 0. Let Z ⊂ Q be a general
union of y tangent vectors. Then h^{0}(I_{Z}(a, b)) = max{0, (a + 1)(b + 1) − 2y} and
h^{1}(I_{Z}(a, b)) = max{0, 2y − (a + 1)(b + 1)}.

Proof. By the semicontinuity theorem for cohomology ([5], III.12.8) it is suffi-
cient to find a disjoint union W ⊂ Q of y tangent vectors such that h^{0}(IW(a, b)) =
max{0, (a + 1)(b + 1) − 2y}. It is obviously sufficient to do it for the integers

y = b(a+1)(b+1)c and y = d(a+1)(b+1)/2e. First assume a odd. Let L_{0}, . . . , L_{b}
be b+1 distinct lines of type (0, 1). Let E_{i}⊂ L_{i}be any disjoint union of (a+1)/2
tangent vectors. In this case we may take W = E1∪ · · · ∪ E_{b}. In the same way
we conclude if b is odd. Hence we may assume that both a and b are even. If
b = 0, then take y tangent vectors of L_{0}. Similarly we conclude if a = 0. Hence
we may assume a ≥ 2 and b ≥ 2 and use induction on a. It is obviously suffi-
cient to check the integers y such that 2y ≥ (a + 1)(b + 1) − 1. Fix a smooth
C ∈ |O_{Q}(2, 2)|. C is a smooth elliptic curve and in particular it is irreducible.

Take a general S ⊂ C with ](S) = a+b. Let W ⊂ C be the union of the 2-points
of C with the points of S as their support, i.e. the degree 2a+2b effective divisor
of C in which each point of S appears with multiplicity two. Let W^{0} ⊂ Q be a
union of y − a − b general tangent vectors. Set Z := W ∪ W^{0}. By the inductive
assumption we have h^{0}(I_{W}^{0}(a − 2, b − 2)) = max{0, (a − 1)(b − 1) − 2y + 2a + 2b},
i.e. h^{0}(I_{W}^{0}(a − 2, b − 2)) = max{0, (a + 1)(b + 1) − 2y} and h^{1}(I_{W}^{0}(a − 2, b − 2)) =
max{0, 2y − (a + 1)(b + 1)}. There are only finitely many (four in characteris-
tic 6= 2, one or two in characteristic 2) line bundles R with R^{⊗2} ∼= OC(a, b).

Since C has genus > 0 for general S the line bundle O_{C}(S) is not one of them.

Hence W /∈ |O_{C}(a, b)|. Since deg(W ) = deg(OC(a, b)), Riemann-Roch gives
h^{i}(C, O_{C}(a, b)(−W )) = 0, i = 0, 1. Since Res_{C}(Z) = W^{0}, the Castelnuovo’s
sequence gives h^{i}(I_{Z}(a, b)) = h^{i}(I_{W}^{0}(a − 2, b − 2)). ^{QED}

We have u3,k := d(k + 3)(k + 1)/6e and v3,k := (k + 2)u3,k − ^{3+k}_{3} . Write
k = 6m + b with 0 ≤ b ≤ 5. We have u_{3,6m} = 6m^{2}+ 4m + 1, v_{3,6m} = 3m + 1,
u_{3,6m+1}= 6m^{2}+ 6m + 2, v_{3,6m+1}= 4m + 2, u_{3,6m+2}= 6m^{2}+ 8m + 3, v_{3,6m+2}=
3m + 2, u3,6m+3 = 6m^{2}+ 10m + 4, v3,6m+3 = 0, u3,6m+4 = 6m^{2} + 12m + 6,
v3,6m+4 = m + 1, u3,6m+5 = 6m^{2} + 14m + 8, v3,6m+5 = 0. The construction
below works (in particular Lemma 6) only because u_{3,6m+7}−v_{3,6m+7}≥ u_{3,6m+5}−
v3,6m+5 (both sides of the inequality are equal to u3,6m+5 = 6m^{2} + 14m + 8).

Without this inequality we would have needed a longer proof. In general we
need u_{3,k+2}− v_{3,k+2}≥ u_{3,k}− v_{3,k} for all k > 0, but only in the case k = 6m + 5
the right hand side is not much bigger than the left hand side.

Let Q ⊂ P^{3} be a smooth quadric surface. We have Pic(Q) ∼= Z^{2} and we take
two distinct, but intersecting, lines contained in Q as a basis of Pic(Q). We will
call |OQ(1, 0)| and |OQ(0, 1)| the two rulings of Q and call any D ∈ |OQ(a, b)| a
divisor (or a curve) of type (a, b). Fix a line L ⊂ Q, P ∈ L and let γ be the set of
all +lines A ⊂ P^{3} with L as their support and P as the support of the nilradical
sheaf of OA. The set γ is the complement of a point in a two-dimensional
projective space (it is P(TPP^{3}) \ P(TPL)). The set γ^{0} of all A ∈ γ contained
in Q is the line P(TPQ) minus the point P(TPL). If A ∈ γ^{0}, then A ⊂ Q and
ResQ(A) = ∅. If A /∈ γ^{0}, then A ∩ Q = L (as schemes) and ResQ(A) = {P } (as
schemes). Now take A ⊂ Q, see L as a divisor of Q; we have Res_{L}(A) = {P }.

Lemma 5. A_{r,1} and A_{3,2} are true.

Proof. Ar,1 is is true by Remark 3.

We have v_{3,2}= 2 and u_{3,2}− v_{3,2}= 1. Take any B = L_{1}t L_{2}t L_{3} ∈ L(3, 3, 0).

B is contained in a unique quadric surface, Q^{0}, and Q^{0} is a smooth quadric. Let
A ⊂ P^{3}be a general +line with L3as its support. We have L1∪L_{2}∪A ∈ L(3, 2, 1).

Since A * Q^{0}, we have h^{0}(I_{L}_{1}∪L_{2}∪A(2)) = 0. ^{QED}
Of course, A_{r,k} makes sense only if u_{r,k}− v_{r,k} ≥ 0; this is the reason why
we didn’t defined A_{r,0}. By the semicontinuity theorem for cohomology ([5],
III.12.8) to prove A3,k it is sufficient to find A ∈ L(3, v3,k, u3,k − v_{3,k}) such
that h^{0}(I_{A}(k)) = 0. Fix an integer k > 0 and let X ⊂ P^{3} be a general union
of v_{3,k} lines and u_{3,k}− v_{3,k} +lines. We have h^{0}(O_{X}(k)) = (u_{3,k}− v_{3,k})(k + 2) +
v_{3,k}(k + 1) = ^{k+3}_{3} , the latter equality being true by the definition of the integer
v_{3,k}. Hence h^{1}(IX(k)) = h^{0}(IX(k)).

Lemma 6. A_{3,k} ⇒ A_{3,k+2} for all k > 0.

Proof. We have 0 ≤ u_{3,k+2}− u_{3,k} ≤ k + 2.

(a) First assume v3,k+2 ≥ v_{3,k}, i.e. k ≡ 0, 3, 4, 5 (mod 6). Notice that in all
cases we have u_{3,k}− v_{3,k} ≤ u_{3,k+2}− v_{3,k+2}(we even have equality if k = 6m + 5,
because u_{3,6m+7}= 6m^{2}+ 18m + 14, v_{3,6m+7}= 4m + 6, u_{3,6m+5}= 6m^{2}+ 14m + 8
and v3,6m+5 = 0). Let L1 ⊂ Q be the union of v_{3,k+2}− v_{3,k} distinct lines of
type (1, 0) and L2 ⊂ Q the union of (u_{3,k+2}− v_{3,k+2}) − (u_{3,k} − v_{3,k}) distinct
lines of type (1, 0) with L_{1}∩ L_{2} = ∅. Let A_{2} ⊂ Q be the union of (u_{3,k+2}−
v_{3,k+2})−(u_{3,k}−v_{3,k}) general +lines contained in Q and with the lines of L_{2}as its
support. Let S2⊂ L_{2}be the support of the nilradical of OA2. Take a general Y ∈
L(3, v_{3,k}, u_{3,k}−v_{3,k}). For general Y we have Y ∩(L_{1}∪L_{2}) = ∅ and so Y ∪A_{2}∪L_{1} ∈
L(3, v_{3,k+2}, u_{3,k+2}− v_{3,k+2}). By the semicontinuity theorem for cohomology ([5],
III.12.8) it is sufficient to prove h^{0}(IY ∪A2∪L1(k)) = 0. Since ResQ(Y ∪A2∪L_{1}) =
Y and h^{0}(I_{Y}(k −2)) = 0, it is sufficient to prove h^{0}(Q, I_{Q∩(Y ∪A}_{2}_{∪L}_{1}_{)}(k +2)) = 0.

Since L_{1}∪ L_{2} ⊂ (Y ∩ Q) ∪ A_{2} ∪ L_{1} and L_{1}∪ L_{2} ∈ |O_{Q}(u_{3,k+2}− u_{3,k}, 0)|, it is
sufficient to prove h^{0}(Q, IRes_{L1∪L2}((Y ∩Q)∪A2∪L1)(k +2−u3,k+2+u3,k, k +2)) = 0.

We have ResL1∪L2((Y ∩ Q) ∪ A2∪ L_{1}) = (Y ∩ Q) ∪ S2. For general A2 the set
S_{2} is a set containing a general point of (u_{3,k+2}− v_{3,k+2}) − (u_{3,k}− v_{3,k}) general
lines of type (1, 0) and nothing else. Hence S_{2} may be considered as a general
union of (u3,k+2 − v_{3,k+2}) − (u3,k − v_{3,k}) points of Q. For general Y the set
Y ∩ Q is a general subset of Q with cardinality 2u_{3,k}. Hence it is sufficient
to check that ]((Y ∩ Q) ∪ S_{2}) = h^{0}(Q, O_{Q}(k + 2 − u_{3,k+2}+ u_{3,k}, k + 2)), i.e.

2u3,k + (u3,k+2− v_{3,k+2}) − (u3,k − v_{3,k}) = (k + 3 − u3,k+2 + u3,k)(k + 3), i.e.

2u_{3,k}+ (k + 2)(u_{3,k+2}− u_{3,k}) = (k + 3)^{2}+ v_{3,k+2}− v_{3,k}. Taking the difference of

(4) for the integer k^{0} = k + 2 from (4) and using that ^{k+5}_{3} − ^{k+3}_{3} = (k + 3)^{2}
we get 2u_{3,k} + (k + 2)(u_{3,k+2}− u_{3,k}) = (k + 3)^{2}+ v_{3,k+2}− v_{3,k}, as wanted.

(b) Now assume v_{3,k+2} < v_{3,k}, i.e. k ≡ 1, 2 (mod 6). Take a general Y ∈
L(3, v_{3,k}, u_{3,k}− v_{3,k}) and write Y = E ∪ F with E ∈ L(3, v_{3,k+2}, u_{3,k}− v_{3,k}) and
F ∈ L(3, v3,k−v_{3,k+2}, 0). For general Y we have h^{0}(IY(k)) = 0 (by the inductive
assumption) and Y ∩ Q is a general subset of Q with cardinality 2u_{3,k}. For each
line L ⊆ F fix one of the point P_{L}∈ L∩Q and call v_{L}a general tangent vector of
Q at PL. Let AL= L ∪ vL⊂ P^{3}be the +lines with L as its reduction, PLas the
support of its nilradical and containing v_{L}. Set G := ∪_{L∈F}A_{L}. Let M ⊂ Q be a
union of u_{3,k+2}− u_{3,k} general lines of type (1, 0). Let N ⊂ Q be a general union
of u3,k+2−u_{3,k}+lines with M as the union of their support. We have E ∪G∪N ∈
L(3, v_{3,k+2}, u_{3,k+2}− v_{3,k+2}). Since Res_{Q}(E ∪ G ∪ N ) = Y and h^{0}(I_{Y}(k)) = 0, it is
sufficient to prove h^{0}(Q, I_{Q∩(E∪G∪N )}(k + 2, k + 2)). Since M ⊂ Q ∩ (E ∪ G ∪ N ),
it is sufficient to prove h^{0}(Q, IResM(Q∩(E∪G∪N ))(k +2−u3,k+2+u3,k, k +2)) = 0.

The scheme G ∩ Q is a general union δ of v_{3,k} tangent vectors of Q and a general
union of v_{3,k} − v_{3,k+2} points of Q; we do not want to use here that general
tangent vectors gives the maximal possible number of conditions to any linear
system, because it requires characteristic zero ([4], [1], Lemma 1.4); however,
since v_{3,k} ≤ 2(k+2)/3, it is obvious that h^{1}(Q, I_{δ}(k+2−u_{3,k+2}+u_{3,k}, k+2)) = 0;

alternatively, use Lemma 4. Set S := ResM(N ). The set S contains one point for
each line of M and it is general with this condition. Since M is a general union
of u_{3,k+2}− u_{3,k} lines of type (1, 0), S may be considered as a general subset of
Q with its cardinality. The set E ∩ Q is a general subset of Q with cardinality
2u_{3,k}− 2v_{3,k+2}. Since 2(v_{3,k}− v_{3,k+2}) + (v_{3,k}− v_{3,k−2}) + (u_{3,k+2}− u_{3,k}) + 2(u_{3,k}−
v_{3,k+2}) = (k + 3 − u_{3,k+2}+ u_{3,k})(k + 3), we are done. ^{QED}

Lemma 7. For all integers k > 0 and c > 0 such that c(k + 2) ≤ ^{k+3}_{3} we
have h^{1}(I_{X}(k)) = 0 for a general X ∈ L(3, 0, c).

Proof. If k = 1, then c = 1. The lemma is obvious in this case.

Now assume k = 2. It is sufficient to do the case c = 2. Take A = A1∪ A_{2} ∈
L(3, 0, 2) with L_{1} and L_{2} two different lines of type (1, 0) of Q, A_{1} ⊂ Q and
general with these restrictions, A_{2} * Q and general among the +lines supported
by Q. We get h^{0}(Q, OQ∩A(2)) = 2 and h^{0}(IResQ(A)) = 0, because ResQ(A) 6= ∅.

From now on we assume k ≥ 3. Lemmas 5 and 6 give that A_{3,k−2} and A_{3,k}
are true. We have c ≤ u_{3,k} and c ≤ u_{3,k} − 1 if v_{3,k} > 0. If c ≤ u_{3,k} − v_{3,k},
then we may use A3,k. In particular we are done if v3,k = 0. Hence we may
assume v_{3,k} > 0. In this case it is sufficient to do the case c = u_{3,k} − 1. Fix a
general Y ∈ L(3, v_{3,k−2}, u_{3,k−2}− v_{3,k−2}). We have h^{i}(I_{Y}(k − 2)) = 0, i = 0, 1,
by A3,k−2. We mimic part (b) of the proof of Lemma 6. Write Y = E t F
with E ∈ L(3, 0, u_{3,k} − v_{3,k}) and F ∈ L(3, v_{3,k}, 0). For each line L ⊆ F fix

one of the point P_{L} ∈ L ∩ Q and call v_{L} a general tangent vector of Q at
P_{L}. Let A_{L} = L ∪ v_{L} ⊂ P^{3} be the +line with L as its reduction, P_{L} as the
support of its nilradical and containing vL. Set G := ∪L∈FAL. Let M ⊂ Q be
a union of u_{3,k+2}− u_{3,k}− 1 general lines of type (1, 0). Let N ⊂ Q be a general
union of u_{3,k+2} − u_{3,k} − 1 +lines with M as the union of their support. Take
X := E ∪ G ∪ N ∈ L(3, 0, u3,k− 1). Since v_{3,k} ≤ k + 1, as in part (b) of the proof
of Lemma 6 we get h^{1}(Q, I_{X∩Q}(k)) = 0 and hence h^{1}(I_{X}(k)) = 0. ^{QED}

Proof of Theorem 1 for r = 3: It is sufficient to prove H_{3,k} for all k ≥ 2
(Remark 3). Fix an integer k > 0. It is sufficient to check the Hilbert function in
degree k of a general element of L(3, t, c) with either t = 0 and (k + 2)c ≤ ^{k+3}_{3}
or (k + 1)t + (k + 2)c = ^{k+3}_{3}

(Remark 4). By Lemma 7 it is sufficient to
check the pairs (t, c) with t > 0 and (k + 1)t + (k + 2)c = ^{k+3}_{3} , i.e. (since the
integers k + 1 and k + 2 are coprime) the pairs (t, c) with t = v_{3,k} + (k + 2)α,
c = u_{3,k}− v_{3,k}− (k + 1)α for some non-negative integer α such that (k + 1)α ≤
u3,k− v_{3,k}. By [6] we may assume c > 0. By Lemma 6 we may assume t > v3,k.
Since v3,2= 2, u3,2= 3 and A3,2 is true (Lemma 5), we may assume k ≥ 3. By
induction on k we may assume that h^{i}(I_{W}(k − 2)) = 0, i = 0, 1, for a general
W ∈ L(3, t^{0}, c^{0}) for all non-negative integers t^{0}, c^{0}such that (k−1)t^{0}+kc^{0} = ^{k+1}_{3} .

Fix a general Y ∈ L(3, m_{3,k−2}− n_{3,k−2}, n_{3,k−2}). We have h^{i}(IY(k − 2)) = 0,
i = 0, 1.

Claim 1: We have t + c ≥ m_{3,k−2}.

Proof of Claim 1: Assume t + c ≤ m_{3,k−2}− 1, i.e. assume (k − 1)(t + c) +
k − 1 ≤ ^{k+1}_{3} . Since (k + 1)t + (k + 2)c = ^{k+3}_{3} , we get (k + 2)/(k − 1) >

k+3

3 / ^{k+1}_{3} = (k + 3)(k + 2)/k(k − 1), a contradiction.

Claim 2: We have c ≥ n_{3,k−2}.

Proof of Claim 2: If k − 2 ≡ 0, 1 (mod 3), then n_{3,k−2} = 0. If k − 2 ≡ 2
(mod 3), then n_{3,k−2}= (k − 1)/3. Hence we may assume k ≡ 4 (mod 3). Since
n_{3,k} = 0, c > 0 and (k + 1)t + (k + 2)c = ^{k+3}_{3} , we get c = β(k + 1) for some
integer β > 0. Hence c ≥ k + 1 > n_{3,k−2}.

Notice that m_{3,k−2} ≥ n_{3,k−2}. Fix a general Y ∈ L(3, m_{3,k−2}−n_{3,k−2}, n_{3,k−2}).

By the inductive assumption we have h^{i}(IY(k − 2)) = 0, i = 0, 1. Set e :=

t + c − m_{3,k−2}. Claim 1 gives e ≥ 0. Take a general union M ⊂ Q of e lines of
type (1, 0).

(a) Assume c − n_{3,k−2} ≤ e. Claim 2 gives c − n_{3,k−2} ≥ 0. Write M =
M1t M_{2} with M2 a union of c − n3,k−2 lines and M1 a union of e − (c − n3,k−2)
lines. Let A_{2} ⊂ Q be a general union of c − n_{3,k−2} +lines with the lines in
M_{2} as their support. Let S_{2} be the support of the nilpotent sheaf of A_{2}. Since
A2 is general, S2 is obtained taking for each line L ⊆ M2 a general point of
L. Set X := Y ∪ M_{1}∪ A_{2}. Since X ∈ L(3, t, c), it is sufficient to prove that

h^{1}(I_{X}(k)) = 0. Since Res_{Q}(X) = Y and h^{1}(I_{Y}(k − 2)) = 0, it is sufficient to
prove that h^{1}(Q, I_{X∩Q}(k)) = 0. Since Res_{M}(X ∩Q) = (Y ∩Q)∪S_{2}, it is sufficient
to prove that h^{1}(Q, I_{(Y ∩Q)∪S}_{2}(k − e, k)) = 0. Since M2 is general and for each
line L ⊂ M_{2}the set S_{2}∩L is a general point of L, S_{2}is a general subset of Q with
cardinality c − n_{3,k−2}. Since Y is general, the set (Y ∩ Q) ∪ S_{2} is a general subset
of Q of cardinality 2m_{3,k−2}+ c − n_{3,k−2}. Since (k − 1)m_{3,k−2}+ n_{3,k−2} = ^{k+1}_{3} ,
(k + 1)t + (k + 2)c = ^{k+3}_{3} , ^{k+3}_{3} − ^{k+1}_{3} = (k + 1)^{2} and e = t + c − m_{3,k−2},
we have 2m3,k−2+ (k + 1)e + c − n3,k−2 = (k + 1)^{2}, i.e. ](S2 ∪ (Y ∩ Q)) =
(k + 1)(k + 1 − e) = h^{0}(Q, O_{Q}(k − e, k)). Hence h^{i}(Q, I_{(Y ∩Q)∪S}_{2}(k − e, k)) = 0,
i = 0, 1.

(b) Now assume c − n_{3,k−2} > e. For each line R ⊆ M fix a general
OR ∈ R and call v_{R} a general tangent vector of Q with OR as its support.

Set R^{+} := R ∪ vR, M^{+} := ∪R⊆MR^{+} and S := ∪R⊆MOR. Since M is general
and each O_{R} is general in R, S is a general subset of Q with cardinality e.

We have M^{+} ⊂ Q and hence M^{+}∩ Q = M^{+} and ResQ(M^{+}) = ∅. Set g :=

c − n_{3,k−2}− e. Since e = t + c − m_{3,k−2}, we get m_{3,k−2}− n_{3,k−2}= g + t > t. Write
Y = Y_{1} t Y_{2} with Y_{2} ∈ L(3, 0, n_{3,k−2}) and Y_{1} ∈ L(3, m_{3,k−2}− n_{3,k−2}, 0). Since
m3,k−2− n_{3,k−2}= g + t ≥ t, we may write Y1 = Y3t Y_{4} with Y3 ∈ L(3, t, 0) and
Y4 ∈ L(3, g, 0). For each line L ⊆ Y_{4} fix one of the two points, say OL, of L ∩ Q
and let w_{L} be a general tangent vector of Q with O_{L} as its support; set L^{+}:=

L ∪ w_{L}∈ L(3, 0, 1). Set Y_{4}^{+}:= ∪_{L⊆Y}_{4}L^{+} and X^{0} := M^{+}∪ Y_{4}^{+}∪ Y_{2}∪ Y_{3}. Since
X^{0} ∈ L(3, t, c), it is sufficient to prove that h^{1}(I_{X}^{0}(k)) = 0. Since ResQ(X^{0}) = Y
and h^{1}(I_{Y}(k−2)) = 0, by the Castelnuovo’s sequence it is sufficient to prove that
h^{1}(Q, I_{X}^{0}_{∩Q}(k)) = 0. The scheme X^{0}∩ Q is the union of M^{+}and (Y_{3}∪ Y_{4}^{+}) ∩ Q.

We have ResM(M^{+}∪ (Y_{3}∪ Y_{4}^{+}) ∩ Q) = S ∪ ((Y \ Y4) ∩ Q) ∪S

L⊆Y4wL. Since
Y ∩ Q is a general subset of Q with cardinality 2m_{3,k−2}, the scheme Z :=

S ∪ ((Y \ Y_{4}) ∩ Q) ∪S

L⊆Y4w_{L} is a general union of g tangent vectors of Q and
e + 2m3,k−2− g points of Q. By the Castelnuovo’s sequence it is sufficient to
prove that h^{1}(Q, I_{Z}(k − e, k)) = 0. By Lemma 4 it is sufficient to check that
deg(Z) ≤ (k − e + 1)(k + 1). By (3) for the integers r = 3 and k^{0}= k − 2 and the
equality (k + 1)t + (k + 2)c = ^{k+3}_{3} we get deg(Z) = (k + 1)(k − e + 1). ^{QED}

3 When r > 3

In this section we prove Theorem 1 for all integers r ≥ 4. For numerical reasons this is easier than in the case r = 3 (as it was in [6] and [2]). The proof in characteristic zero is very short and we will only give it.

Lemma 8. For all integers r ≥ 3 and k ≥ 2 we have m_{r,k−1}< u_{r,k}.

Proof. We have km_{r,k−1}≤ ^{k+r−1}_{r} and (k + 2)u_{r,k}≥ ^{r+k}_{r} . Note that

r + k r

.r + k − 1 r

= (r + k)/k

and that (r + k)k > (k + 2)k for all r ≥ 3. ^{QED}
We need the following assumption B_{r,k}:

Br,k, r ≥ 4, k > 0. Fix a hyperplane H ⊂ P^{r}. There is X ∈ L(r, mr,k−
n_{r,k}, n_{r,k}) such that the support of the nilradical sheaf of X is contained in H
and h^{0}(I_{X}(k)) = 0.

For all X ∈ L(r, m_{r,k} − n_{r,k}, n_{r,k}) we have h^{0}(O_{X}(k)) = ^{r+k}_{r}

and so
h^{1}(I_{X}(k)) = h^{0}(I_{X}(k)).

Lemma 9. For all integers r ≥ 4 and k ≥ 2 we have mr,k≥ m_{r,k−1}.
Proof. We have

m_{r,k−1}+ (k + 1)(m_{r,k}− m_{r,k−1}) + n_{r,k}− n_{r,k−1} =r + k − 1
r − 1

(5)
Assume m_{r,k}≤ m_{r,k−1}− 1. Since n_{r,k}− n_{r,k−1}≤ k, (6) gives

m_{r,k−1}− 1 ≥r + k − 1
r − 1

Since km_{r,k−1} ≤ ^{r+k−1}_{r} and k ^{r+k−1}_{r−1} = r ^{r+k−1}_{r} , we get −k ≥ (r − 1) ^{r+k−1}_{r−1} ,

a contradiction. ^{QED}

Lemma 10. Fix an integer r ≥ 4 and assume that Theorem 1 is true in
P^{r−1}. Then B_{r,k} is true for all k > 0.

Proof. Br,1is true by Remark 3. Hence we may assume k ≥ 2 and that B_{r,k−1}is
true. Fix Y ∈ L(r, m_{r,k−1}−n_{r,k−1}, n_{r,k−1}) such that the support of the nilradical
sheaf of Y is contained in H and h^{0}(I_{Y}(k − 1)) = 0. By the semicontinuity
theorem for cohomology ([5], III.12.8) we may assume that Y is general among
the elements of L(r, m_{r,k−1}− n_{r,k−1}, n_{r,k−1}) whose nilradical sheaf is supported
by points of H. Hence we may assume that no irreducible component of Y_{red}
is contained in H, that Yred is a general subset of H with cardinality mr,k−1

and that for each +line A ⊂ Y , say A = L ∪ v_{L}, the tangent vector v_{L} of A is
not contained in H. The latter assumption implies Y_{red}∩ H = Y ∩ H (scheme-
theoretic intersection) and ResH(Y ) = Y . We have mr,k ≥ m_{r,k−1} (Lemma
9).

(a) In this step we assume n_{r,k} < n_{r,k−1}. Let F ⊂ H be a general union
of m_{r,k} − m_{r,k−1} lines, with the only restriction that exactly n_{r,k−1}− n_{r,k} of
them contain a point of Y_{red}∩ H. We have m_{r,k−1}− n_{r,k−1}− (n_{r,k−1}− n_{r,k}) +
m_{r,k}− m_{r,k−1} = m_{r,k}− 2n_{r,k−1} + n_{r,k}. The scheme Y ∪ F is a disjoint union
of n_{r,k−1}− n_{r,k} sundials, n_{r,k} +lines and m_{r,k} − 2n_{r,k−1} + n_{r,k} lines. Since a
sundial is a flat limit of a family of elements of L(r, 2, 0) ([2]), it is sufficient
to prove h^{0}(I_{Y ∪F}(k)) = 0. Since the set Y_{red}∩ H is general in H, F may be
considered as a general union of lines. Hence F has maximal rank. By (5) we
have h^{1}(H, IF(k)) = 0 and h^{0}(H, IF(k)) = mr,k−1 − n_{r,k−1} + nr,k. Since for
fixed Y_{red}∩ H ∩ F we may deform the other components of Y so that the
other m_{r,k−1} − n_{r,k−1} + n_{r,k} points of Y_{red}∩ (H \ F ) are general in H, then
h^{i}(H, I_{H∩(Y ∪F )}(k)) = 0. Castelnuovo’s sequence gives h^{0}(IY ∪F(k)) = 0.

(b) In this step we assume n_{r,k} ≥ n_{r,k−1} and m_{r,k}− n_{r,k} ≥ m_{r,k−1} −
n_{r,k−1}. Let E ⊂ H be a general union of m_{r,k}− n_{r,k}− (m_{r,k−1}− n_{r,k−1}) lines
and n_{r,k} − n_{r,k−1} +lines. We have Y ∪ E ∈ L(r, m_{r,k} − n_{r,k}, n_{r,k}) and the
support of the nilradical sheaf of Y ∪ E is contained in H. By (5) we have
h^{0}(E, OE(k)) + deg(Y ∩ H) = ^{r+k−1}_{r−1} . Since Theorem 1 is true in P^{r−1}, we
have h^{1}(H.IE(k)) = 0. Since Y ∩ H is a general union of m_{r,k−1} points of H,
(5) implies h^{i}(H, I_{(Y ∪E)∩H}(k)) = 0.

(c) In this step we assume n_{r,k}≥ n_{r,k−1}and m_{r,k}− n_{r,k}< m_{r,k−1}− n_{r,k−1}.
Therefore g := n_{r,k}−n_{r,k−1}−(m_{r,k}−m_{r,k−1}) > 0. Since n_{r,k}≤ k, we have g ≤ k.

Since km_{r,k−1}+n_{r,k−1} = ^{r+k−1}_{r} and n_{r,k−1}≤ k−1, we have g ≤ m_{r,k−1}−n_{r,k−1}.
Take a general union G ⊂ H of mr,k−m_{r,k−1}+lines. Since Theorem 1 is assumed
to be true in P^{r−1}, G has maximal rank. By (5) we have h^{1}(H, I_{G}(k)) = 0 and
h^{0}(H, I_{G}(k)) = m_{r,k−1}+ g. Write Y = Y_{1}t Y_{2}t Y_{3} with Y_{3} ∈ L(r, 0, n_{r,k−1}),
Y1 ∈ L(r, m_{r,k−1}−n_{r,k−1}, 0) and Y2 ∈ L(r, g, 0). For each line L ⊆ Y_{2}let vLbe the
general tangent vector of H with L ∩ H as its support. Set A2 := ∪L⊆Y2(L ∪ vL).

Since U := Y_{1}∪ A_{2}∪ Y_{3}∪ G ∈ L(r, m_{r,k}− n_{r,k}, n_{r,k}), it is sufficient to prove that
h^{0}(IU(k)) = 0. We have ResH(U ) = Y , because each vLis contained in H. The
scheme U ∩ H is the union of G, m_{r,k−1}− n_{r,k−1}− g general points of H and g
general tangent vectors of H. Hence h^{i}(H, I_{U ∩H}(k)) = 0, i = 0, 1 ([1], Lemma

1.4). ^{QED}

Proof of Theorem 1 for r > 3: Let H ⊂ P^{r} be a hyperplane. We use
induction on r, the starting case being the one with r = 3 proved in section 2.

Hence we assume Theorem 1 in H ∼= P^{r−1} for all L(r − 1, t^{0}, c^{0}). By Remark 3 it
is sufficient to prove H_{r,k} for all k > 0. Hr,1 is true (Remark 3). Hence we may
assume k ≥ 2 and that H_{r,k−1} is true. By Remark 4 it is sufficient to prove H_{r,k}
for the pairs (t, c) such that either t = 0 and ^{r+k}_{r} − k − 1 ≤ c(k + 2) ≤ ^{r+k}_{r}
or t(k + 1) + (k + 2)c = ^{r+k}_{r}

and c > 0. If ^{r+k}_{r} − k − 1 ≤ c(k + 2) ≤